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The NBA's Least Remarkable Stat

May 19, 2005 11:00 AM
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I've read in a few places now that Game 5s of a 2-2 series are pivotal ... the winner has taken the series ... 84 percent.

That sounds like quite an advantage — but there's no momentum, home court advantage or anything else at work. Those are basically exactly the numbers you'd expect.

Let me explain:

Let's say that the Detroit Pistons are beating the Indiana Pacers 3-2. Based on their results in this series, and ignoring any other factors, we'd say that Detroit has a 60% chance of winning the next game against the Pacers.

So, we could estimate that the odds of Indy winning the next two games are 40% times 40%, or 16%. Ergo, the odds of the Pistons winning at least one of the two remaining games are 100% minus 16%, or exactly the 84% cited in the article.

If we assume that on the 40% chance of a Pacers victory in game six, they have a 50/50 probability of winning game seven, the odds of them winning both are 40% times 50%, or 20%. Ergo, the odds should be 100% minus 20%, or 80%, that a team with a 3-2 lead will end up winning the series.

So, at best, things like momentum, home court advantage, a better regular season record, player injuries, etc. barely matter when the series is at 3-2; their net impact has a mere 4% effect on the outcome.


Comments: The NBA's Least Remarkable Stat

Interesting.

Posted by: Bubba on May 19, 2005 4:49 PM | permalink

Wait, let's think about this for a minute.

Indy's chance of winning assuming the probabilities are based only on the outcomes of the previous games is .4 *.5 = .2, or 20%.

By the same logic, Detroit's chances of winning in a 7th game is .6 *.5 = .3, or 30%.

The two of these together make only 50%. The same logic also dictates a 0% chance for one team winning after game 1 and a 0% chance for either team winning before the series starts.

Posted by: ChuckJerry on May 19, 2005 7:11 PM | permalink

Jerry, you're wrong on one point and right on the other.

1) "The two of these together make only 50%."

That's because you're calculating the wrong odds for a 4-3 Pistons win and omitting a third possibility: Detroit winning after 6.

Here are the odds of each scenario, by this 3-2 extrapolation only:

Detroit wins in 6: 60%
Detroit wins in 7: 20% (40% chance of losing game six, with a 50% chance on winning game 7) -- you mistakenly did .6 x .5 on this one.
Indy wins in 7: 20% (40% chance of winning game six, with a 50% chance on winning game 7)

It totals up to 100%.

2) What you are right about is that it's wrong to say a 1-0, 2-0 or even 3-0 lead means 100% victory.

However, despite the divide-by-zero problem (if you have won 0 of 0 games in the series, your percentage is 0/0 * 100%) ... it's a tautology that the *average* team has a 50/50% chance of winning when the series starts, or when it's tied at 1-1, 2-2 or 3-3. Think about it for a minute: if there were, say 120 series that went to a 3-3 tie, you have 60 winners and 60 losers. Therefore, the average team tied up, going into a game seven, wins half the time. But this is meaningless.

Your example of a 1-0 lead, though, does indeed show that extrapolating makes less and less sense when the sample is smaller (example: the number of 3-0 teams who emerge victorious is a lot closer to 100% than the number of 1-0 squads that do).

I'd be interested, though, what the numbers are for 3-1 series leads. By this reckoning, the odds of victory are 95% (100% - (25% * 40% * 50%)).

Posted by: Joe Grossberg on May 19, 2005 9:14 PM | permalink

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